TITLE.PM5

762 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th14-4.pm5

p (Pressure)

h (Enthalpy)

(^51315971790) kJ/kg
(^12) – 12°C
Evaporation
Condensation

• 27°C
  21°C 27°C
  4
  Sub-cooling
  in condenser
  3
  ThrottlingThrottling CompressionCompression
  Constant
  entropy line
  Constant
  entropy line
  Fig. 14.36
  (i)Condition of the vapour at the outlet of the compressor
  = 58 – 27 = 31 °C superheat. (Ans.)
  (ii)Condition of vapour at entrance to evaporator,
  x 1 = 0.13. (Ans.)
  (iii) C.O.P. =
  hh
  hh
  21
  32
  1597 513
  1790 1597
  −
  −
  = −
  − = 5.6. (Ans.)
  (iv)Power required :
  C.O.P. =
  Net refrigerating effect
  Work done
  =R
  W
  n
  5.6 =
  10 5 14000
  60

. ×
W×

∴ W =

10 5 14000

56 60

.

.

×
×
kJ/min = 437.5 kJ/min.
= 7.29 kJ/s.
i.e., Power required = 7.29 kW. (Ans.)

+Example 14.27. The evaporator and condenser temperatures of 20 tonnes capacity freezer
are – 28°C and 23°C respectively. The refrigerant – 22 is subcooled by 3°C before it enters the
expansion valve and is superheated to 8°C before leaving the evaporator. The compression is
isentropic. A six-cylinder single-acting compressor with stroke equal to bore running at 250 r.p.m.
is used. Determine :
(i)Refrigerating effect/kg.
(ii)Mass of refrigerant to be circulated per minute.
(iii)Theoretical piston displacement per minute.
(iv)Theoretical power.